Assume the vehicle speed on a freeway has a bell-shaped distribution with mean s
ID: 3202295 • Letter: A
Question
Assume the vehicle speed on a freeway has a bell-shaped distribution with mean speed of 82 mph and standard deviation of 7.4 mph.
7 If you repeatedly clock 16 vehicles at a time and obtain the mean of each sample, the fraction of sample means that would fall within ±3 mph from the mean speed of all vehicles is ______. a 0.7919 b 0.8249 c 0.8593 d 0.8951 8 With the same sample size as in the previous problem, the middle interval which contains 95% of the mean vehicle speed of all such samples is _____. a 81.1 82.9 b 80.2 83.8 c 79.3 84.7 d 78.4 85.6 9 Assume the population mean price per gallon of gasoline is = $2.35, and the population standard deviation is = $0.10. Suppose that a random sample of 30 gasoline stations will be selected. What is the probability that the simple random sample will provide a sample mean within 3¢ ($0.03) of the population mean? a 0.9316 b 0.9108 c 0.8997 d 0.8729Explanation / Answer
Result:
Assume the vehicle speed on a freeway has a bell-shaped distribution with mean speed of 82 mph and standard deviation of 7.4 mph.
7 If you repeatedly clock 16 vehicles at a time and obtain the mean of each sample, the fraction of sample means that would fall within ±3 mph from the mean speed of all vehicles is ______.
a 0.7919
b 0.8249
c 0.8593
answer: d 0.8951
standard error = sd/sqrt(n) = 7.4/sqrt(16) =1.85
z =3/1.85= 1.62
P( mean ±3 mph) =P( -1.62<z<1.62) = P( z <1.62) – P( z< -1.62)
=0.9474-0.0526 = 0.8948
8 With the same sample size as in the previous problem, the middle interval which contains 95% of the mean vehicle speed of all such samples is _____.
a 81.1 82.9
b 80.2 83.8
c 79.3 84.7
answer: d 78.4 85.6
within 2 standard error , 95% of the mean falls.
lower level = 82-2 *1.85 = 78.3
upper level = 82+2 *1.85 = 85.7
9 Assume the population mean price per gallon of gasoline is = $2.35, and the population standard deviation is = $0.10. Suppose that a random sample of 30 gasoline stations will be selected. What is the probability that the simple random sample will provide a sample mean within 3¢ ($0.03) of the population mean?
standard error = sd/sqrt(n) = 0.10/sqrt(30) =0.0183
z value for within 0.03, z= 0.03/0.0183 =1.64
P( mean ±0.03) =P( -1.64<z<1.64) = P( z <1.64) – P( z< -1.64)
= 0.9495- 0.0505 =0.8990
a 0.9316
b 0.9108
answer: c 0.8997
d 0.8729